For the function `f(x)=x^(2)-6x+8, 2 le x le 4 `, the value of x for which f'(x) vanishes is

Options

A. 3
B. `(5)/(2)`
C. `(9)/(4)`
D. `(7)/(2)`

Answer

To solve the problem, we need to find the value of \( x \) for which the derivative \( f'(x) \) of the function \( f(x) = x^2 - 6x + 8 \) vanishes (i.e., equals zero) within the interval \( [2, 4] \).

### Step-by-Step Solution:

1. **Find the derivative of the function**:
\[ f(x) = x^2 - 6x + 8 \]
To find \( f'(x) \), we differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}(6x) + \frac{d}{dx}(8) \]
\[ f'(x) = 2x - 6 \]

2. **Set the derivative equal to zero**:
We need to find the points where \( f'(x) = 0 \):
\[ 2x - 6 = 0 \]

3. **Solve for \( x \)**:
Rearranging the equation gives:
\[ 2x = 6 \]
\[ x = \frac{6}{2} = 3 \]

4. **Check if \( x = 3 \) is within the given interval**:
The problem states that \( x \) must be in the range \( [2, 4] \). Since \( 3 \) is within this interval, we conclude that:
\[ x = 3 \text{ is the required value.} \]

### Final Answer:
The value of \( x \) for which \( f'(x) \) vanishes is \( \boxed{3} \).